Question

A satellite must move in the equatorial plane of the Earth close to its surface either in the Earth's rotation direction or against it. Find how many times the kinetic energy of the satellite in the latter case exceeds that in the former case (in the reference frame fixed to the Earth).

• A. $1.27$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is A $1.27$

From the well known relationship between the velocities of a particle w.r.t. a space fixed frame ($K$) rotating frame ($K^{\prime }$) $\overrightarrow{v}={\overrightarrow{v}}^{\prime }+(\overrightarrow{w}+\overrightarrow{r})$

$v_1^{\prime }=v-\left(\frac{2\pi }{T}\right)R$

Thus kinetic energy of the satellite in the earth's frame

$T_1^{\prime }=\frac{1}{2}m{v^{\prime }}_1^2=\frac{1}{2}m{(v-\frac{2\pi R}{T})}^2$ .....$\left(1\right)$

Obviously when the satellite moves in opposite sense compared to the rotation of the Earth its velocity relative to the same frame would be

$v_2^{\prime }=v+\left(\frac{2\pi }{T}\right)R$

And kinetic energy

$T_2^{\prime }=\frac{1}{2}m{v^{\prime }}_2^2=\frac{1}{2}m{(v+\frac{2\pi R}{T})}^2$ ......$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$T^{\prime }=\frac{{(v+\frac{2\pi R}{T})}^2}{{(v-\frac{2\pi R}{T})}^2}$ .....$\left(3\right)$

Now from Newton's second law

$\frac{\gamma Mm}{R^2}=\frac{m{v}^2}{R}$ or $v=\sqrt{\frac{\gamma M}{R}}=\sqrt{gR}$ .....$\left(4\right)$

Using $\left(4\right)$ and $\left(3\right)$

$\frac{T_2^{\prime }}{T_1^{\prime }}=\frac{{(\sqrt{gR}+\frac{2\pi R}{T})}^2}{{(\sqrt{gR}-\frac{2\pi R}{T})}^2}=1.27$ nearly (Using appendices)